A class consist of a no. of boys whose ages are in A.P.,the common difference being 4 months.If the youngest boy is just 8 years old and the sum of their ages is 168 years,then find the no. of boys in the class.
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★★★★★★★★★★★★★★★Let us first convert all years in months.
8yrs. = 96 months, 168yrs. = 2016 months.
Therefore A.P. = 96,100,104....
Now, a = 96
S = 2016
d = 4
n=?
S = n/2( 2a + (n-1)d)
2016 = n/2( 192 + n-1)4)
4032 = n( 192 + 4n-4)
4032 = 4n2+188n
4n2 +188n - 4032 = 0
n²+47n -1008 = 0
n²+ 63n-16n-1008 =0
n(n+63) -16(n+63)=0
(n-16) (n+63) =0
n=16 and n= -63
therefore number of students = 16
(we neglected -63 as number of students cannot be negative )
★★★★★★★★★★★★★★★
Answer:
By converting in months
AP:96, 100, 104....
Applying Sn= n/2(2a+(n-1)D)
We can find the number of boys
Number of boys are 16